## Section4.4The Chain Rule

Let $h(x)=\sqrt{625-x^2}\text{.}$ The rules stated previously do not allow us to find $h'(x)\text{.}$ However, $h(x)$ is a composition of two functions. Let $f(x)=\sqrt x$ and $g(x)=625-x^2\text{.}$ Then we see that

\begin{equation*} h(x)=(f\circ g)(x)\text{.} \end{equation*}

From our rules we know that $f'(x)=\frac{1}{2}x^{-1/2}$ and $g'(x)=-2x\text{,}$ thus it would be convenient to have a rule which allows us to differentiate $f\circ g$ in terms of $f'$ and $g'\text{.}$ This gives rise to the Chain Rule.

The Chain Rule has a particularly simple expression if we use the Leibniz notation for the derivative. The quantity $f'(g(x))$ is the derivative of $f$ with $x$ replaced by $g\text{;}$ this can be written $df/dg\text{.}$ As usual, $g'(x)=dg/dx\text{.}$ Then the Chain Rule becomes

\begin{equation*} {df\over dx} = {df\over dg}{dg\over dx}\text{.} \end{equation*}

This looks like trivial arithmetic, but it is not: $dg/dx$ is not a fraction, that is, not literal division, but a single symbol that means $g'(x)\text{.}$ Nevertheless, it turns out that what looks like trivial arithmetic, and is therefore easy to remember, is really true.

It will take a bit of practice to make the use of the Chain Rule come naturally—it is more complicated than the earlier differentiation rules we have seen.

###### Example4.43. Chain Rule.

Compute the derivative of $\ds \sqrt{625-x^2}\text{.}$

Solution

We already know that the answer is $\ds -x/\sqrt{625-x^2}\text{,}$ computed directly from the limit. In the context of the Chain Rule, we have $\ds f(x)=\sqrt{x}\text{,}$ $\ds g(x)=625-x^2\text{.}$ We know that $\ds f'(x)=(1/2)x^{-1/2}\text{,}$ so $\ds f'(g(x))= (1/2)(625-x^2)^{-1/2}\text{.}$ Note that this is a two step computation: first compute $f'(x)\text{,}$ then replace $x$ by $g(x)\text{.}$ Since $g'(x)=-2x$ we have

\begin{equation*} f'(g(x))g'(x)={1\over 2\sqrt{625-x^2}}(-2x)={-x\over \sqrt{625-x^2}}\text{.} \end{equation*}
###### Example4.44. Chain Rule.

Compute the derivative of $\ds 1/\sqrt{625-x^2}\text{.}$

Solution

This is a quotient with a constant numerator, so we could use the Quotient Rule, but it is simpler to use the Chain Rule. The function is $\ds (625-x^2)^{-1/2}\text{,}$ the composition of $\ds f(x)=x^{-1/2}$ and $\ds g(x)=625-x^2\text{.}$ We compute $\ds f'(x)=(-1/2)x^{-3/2}$ using the Power Rule, and then

\begin{equation*} f'(g(x))g'(x)={-1\over 2(625-x^2)^{3/2}}(-2x)={x\over (625-x^2)^{3/2}}\text{.} \end{equation*}

Suppose the only data available to you are some points on the graphs of a function $y=f(x)$ and its derivative $f'(x)\text{.}$

###### Example4.45. Chain Rule and Data Points.

Given

\begin{equation*} f(2)=-1 \text{ , } f(-1)=3 \text{ , } f'(2)=4 \text{ , } f'(-1)=-5 \text{ and, } \end{equation*}
\begin{equation*} g(2)=2 \text{ , } g(-1)=-2 \text{ , } g'(-1)=0 \end{equation*}

If possible, find the following derivatives:

1. $\left(f \circ g \right)'(2)$

2. $\left(f \circ f \right)'(2)$

3. $\left(g \circ f \right)'(-1)$

Solution
1. $\left(f \circ g \right)'(2) = f'\left(g(2)\right) \cdot g'(2) = f'(2) \cdot 7 = 4 \cdot 7 = 28$

2. $\left(f \circ f \right)'(2) = f'\left(f(2)\right) \cdot f'(2) = f'(-1) \cdot 4 = (-5) \cdot 4 = -20$

3. $\left(g \circ f \right)'(-1) = g'\left(f(-1)\right) \cdot f'(-1) = g'(3) \cdot (-5)\text{,}$ which cannot be found since we do not know what $g'(-3)$ evaluates to.

###### Example4.46. Investigating $f(g(a))$ and $g(f(a))$.

The graphs of the functions $y=f(x)$ and $y=g(x)$ are given below. Suppose $h(x) = f(g(x))$ and $k(x)=g(f(x))\text{.}$

1. Find $h(4)$ and $k(4)\text{.}$ Are those values the same?

2. Find $h'(4)$ and $k'(4)\text{.}$ Are those values the same?

Solution
1. Since $h(x)=f(g(x))\text{,}$ we have

\begin{equation*} h(4) = f(g(4))\text{.} \end{equation*}

From the graph of $g$ (in red), we read off that $g(4)=2\text{,}$ therefore

\begin{equation*} h(4) = f(g(4)) = f(2)\text{.} \end{equation*}

From the graph of $f$ (in blue), we read off that $f(2)=0\text{,}$ therefore

\begin{equation*} h(4) = f(2) = 0\text{.} \end{equation*}

Since $k(x) = g(f(x))\text{,}$ we similarly read off from the graphs of $f$ and $g$ respectively to get

\begin{equation*} k(4)=g(f(4)) = g(0) = 4\text{.} \end{equation*}

Comparing values, we see that $h(4) \neq k(4)\text{.}$

2. By the Chain Rule, we have

\begin{equation*} h'(x) = f'(g(x)) \cdot g'(x) \text{ and, } \end{equation*}
\begin{equation*} k'(x) = g'(f(x)) \cdot f'(x)\text{.} \end{equation*}

Therefore,

\begin{equation*} h'(4)=f'(g(4))\cdot g'(4)\text{.} \end{equation*}

We already know from (1) that $g(4)=2\text{,}$ and so

\begin{equation*} h'(4)=f'(2)\cdot g'(4)\text{.} \end{equation*}

Now, the slope of the graph of $f$ at $x=2$ is read off as $-2/1 = -2\text{,}$ and the slope of the graph of $g$ at $x=4$ is read off as $-1/1 = -1\text{.}$ We compute

\begin{equation*} h'(4) = f'(2) \cdot g'(4) = (-2)(-1) = 2\text{.} \end{equation*}

Similarly, we compute

\begin{equation*} k'(4) = g'\left(f(4)\right)\cdot f'(4) = g'(0)\cdot f'(4) = (0)(-2/1) = 0\text{.} \end{equation*}

Comparing values, we conclude that $h'(4) \neq k'(4)\text{.}$

Note: In general,

\begin{equation*} \begin{split} f(g(x)) \amp \color{red} \neq g(f(x)) \ \ \text{ and, } \\ \frac{d}{dx} \left(f(g(x))\right) \amp \color{red} \neq \frac{d}{dx} \left(g(f(x))\right) \end{split} \end{equation*}

In practice, of course, you will need to use more than one of the rules we have developed to compute the derivative of a complicated function.

###### Example4.47. Derivative of Quotient.

Compute the derivative of

\begin{equation*} f(x)={x^2-1\over x\sqrt{x^2+1}}\text{.} \end{equation*}
Solution

The “last” operation here is division, so to get started we need to use the Quotient Rule first. This gives

\begin{equation*} \begin{split} f'(x) =\amp {(x^2-1)'x\sqrt{x^2+1}-(x^2-1)(x\sqrt{x^2+1})'\over x^2(x^2+1)}\\ =\amp {2x^2\sqrt{x^2+1}-(x^2-1)(x\sqrt{x^2+1})'\over x^2(x^2+1)}\end{split}\text{.} \end{equation*}

Now we need to compute the derivative of $\ds x\sqrt{x^2+1}\text{.}$ This is a product, so we use the Product Rule:

\begin{equation*} {d\over dx}x\sqrt{x^2+1}=x{d\over dx}\sqrt{x^2+1}+\sqrt{x^2+1}\text{.} \end{equation*}

Finally, we use the Chain Rule:

\begin{equation*} {d\over dx}\sqrt{x^2+1}={d\over dx}(x^2+1)^{1/2}= {1\over 2}(x^2+1)^{-1/2}(2x)={x\over \sqrt{x^2+1}}\text{.} \end{equation*}

And putting it all together:

\begin{equation*} \begin{split} f'(x) =\amp {2x^2\sqrt{x^2+1}-(x^2-1)(x\sqrt{x^2+1})'\over x^2(x^2+1)}\\ =\amp {2x^2\sqrt{x^2+1}-(x^2-1)\left(x{\ds{x\over \sqrt{x^2+1}}} +\sqrt{x^2+1}\right)\over x^2(x^2+1)} \end{split}\text{.} \end{equation*}

This can be simplified of course, but we have done all the calculus, so that only algebra is left.

Using the Chain Rule, the Power Rule, and the Product Rule, it is possible to avoid using the Quotient Rule entirely.

###### Example4.48. Derivative of Quotient without Quotient Rule.

Compute the derivative of $\ds f(x)={x^3\over x^2+1}\text{.}$

Solution

Write $\ds f(x)=x^3(x^2+1)^{-1}\text{,}$ then

\begin{equation*} \begin{split} f'(x) =\amp x^3{d\over dx}(x^2+1)^{-1}+3x^2(x^2+1)^{-1}\\ \ =\amp x^3(-1)(x^2+1)^{-2}(2x)+3x^2(x^2+1)^{-1}\\ =\amp -2x^4(x^2+1)^{-2}+3x^2(x^2+1)^{-1}\\ =\amp {-2x^4\over (x^2+1)^{2}}+{3x^2\over x^2+1}\\ =\amp {-2x^4\over (x^2+1)^{2}}+{3x^2(x^2+1)\over (x^2+1)^{2}}\\ =\amp {-2x^4+3x^4+3x^2\over (x^2+1)^{2}}={x^4+3x^2\over (x^2+1)^{2}} \end{split} \end{equation*}

Note that we already had the derivative on the second line; all the rest is simplification. It is easier to get to this answer by using the Quotient Rule, so there's a trade off: more work for fewer memorized formulas.

###### Example4.49. Chain Rule and Tangent Line.

Find the slope of the tangent line to the graph of the function

\begin{equation*} f(x)=\left(\frac{2x+1}{3x+2}\right)^{3} \end{equation*}

at the point $\left(0,\frac{1}{8}\right)\text{.}$

Solution

The slope of the tangent line to the graph of $f$ at any point is given by $f'(x)\text{.}$ To compute $f'(x)\text{,}$ we use the general Power Rule followed by the Quotient Rule, obtaining,

\begin{equation*} \begin{split} f'(x) \amp = 3\left(\frac{2x+1}{3x+2}\right)^{2}\frac{d}{dx}\left(\frac{2x+1}{3x+2}\right) \\ \amp = 3\left(\frac{2x+1}{3x+2}\right)^{2}\left[\frac{(3x+2)(2) - (2x+1)(3)}{(3x+2)^{2}}\right] \\ \amp = 3\left(\frac{2x+1}{3x+2}\right)^{2}\left[\frac{6x+4-6x-3}{(3x+2)^{2}}\right] \\ \amp = \frac{3(2x+1)^{2}}{(3x+2)^{4}} \end{split} \end{equation*}

In particular, the slope of the tangent line to the graph at $\left(0,\frac{1}{8}\right)$ is given by

\begin{equation*} f'(0) = \frac{3(0+1)^{2}}{(0+2)^{4}} = \frac{3}{16}\text{.} \end{equation*}
###### Example4.50. Chain of Composition.

Compute the derivative of $\ds \sqrt{1+\sqrt{1+\sqrt{x}}}\text{.}$

Solution

Here we have a more complicated chain of compositions, so we use the Chain Rule three times. At the outermost “layer” we have the function $\ds g(x)=1+\sqrt{1+\sqrt{x}}$ plugged into $\ds f(x)=\sqrt{x}\text{,}$ so applying the Chain Rule once gives

\begin{equation*} {d\over dx}\sqrt{1+\sqrt{1+\sqrt{x}}}= {1\over 2}\left(1+\sqrt{1+\sqrt{x}}\right)^{-1/2}{d\over dx} \left(1+\sqrt{1+\sqrt{x}}\right)\text{.} \end{equation*}

Now we need the derivative of $\ds \sqrt{1+\sqrt{x}}\text{.}$ Using the Chain Rule again:

\begin{equation*} {d\over dx}\sqrt{1+\sqrt{x}}={1\over 2}\left(1+\sqrt{x}\right)^{-1/2}{1\over 2}x^{-1/2}\text{.} \end{equation*}

So the original derivative is

\begin{equation*} \begin{split} {d\over dx}\sqrt{1+\sqrt{1+\sqrt{x}}} =\amp {1\over 2}\left(1+\sqrt{1+\sqrt{x}}\right)^{-1/2} {1\over 2}\left(1+\sqrt{x}\right)^{-1/2}{1\over 2}x^{-1/2}\\ =\amp {1\over 8 \sqrt{x}\sqrt{1+\sqrt{x}}\sqrt{1+\sqrt{1+\sqrt{x}}}}\end{split} \end{equation*}

To see the full power of the Chain Rule, we shall consider more complex compositions of functions.

###### Example4.51. Complex Chain of Composition.

Suppose we are given the functions $f(x)=\sqrt[5]{x}$ , $g(x) = \left(x^{2}+3x\right)^{24}\text{,}$ and, $h(x) = \frac{1}{x+2}\text{.}$ Find the first derivative of the following compositions:

1. $\left(f \circ h \circ g \right)(x)$

2. $\left(g \circ f \circ h \right)(x)$

Solution

We will first make some general observations. By applying the Chain Rule twice, we find that the derivative of $\left(f \circ g \circ h \right)(x)$ is

\begin{equation*} \begin{split} \frac{d}{dx} \left(f \circ g \circ h \right)(x) \amp = \frac{d}{dx} \left(f(g(h(x)))\right)\\ \amp = f'(g(h(x)) \cdot \frac{d}{dx} \left(g(h(x))\right) \\ \amp = f'(g(h(x)) \cdot g'(h(x)) \cdot h'(x) . \end{split} \end{equation*}

The derivative of $f$ is

\begin{equation*} f'(x)=\frac{d}{dx} x^{1/5} = \frac{1}{5}x^{-4/5}\text{.} \end{equation*}

The derivative of $g$ is

\begin{equation*} \begin{split} g'(x) \amp = \frac{d}{dx} \left(x^{2}+3x\right)^{24} \\ \amp =24\left(x^{2}+3x\right)^{23} \cdot \frac{d}{dx}\left(x^{2} + 3x\right) \\ \amp = 24\left(x^{2}+3x\right)^{23} \cdot \left(2x + 3\right) \\ \amp = 24 \left(2x + 3\right)\left(x^{2}+3x\right)^{23}. \end{split} \end{equation*}

The derivative of $h$ is

\begin{equation*} \begin{split} h'(x) \amp = \frac{d}{dx}\left(\frac{1}{x+2}\right)\\ \amp = \frac{(x+2)\frac{d}{dx}(1) - 1\frac{d}{dx}(x+2)}{(x+2)^{2}}\\ \amp = \frac{0(x+2) - 1(1)}{(x+2)^{2}} \\ \amp = -\frac{1}{(x+2)^{2}} \cdot \end{split} \end{equation*}
1. The derivative of $(f \circ h \circ g)(x)$ is therefore given by

\begin{equation*} \frac{d}{dx} (f \circ h \circ g)(x) = f'(h(g(x))\cdot h'(g(x)) \cdot g'(x)\text{.} \end{equation*}

We can now use the derivatives of $f\text{,}$ $g$ and $h$ we calculated above and evaluate them at the appropriate inner functions.

\begin{equation*} \begin{split} f'(h(g(x)) \amp = f'\left(h(x^{2}+3x)^{24})\right) \\[1ex] \amp =f'\left(\frac{1}{(x^{2}+3x)^{24}+2}\right)\\[1ex] \amp = \frac{1}{5}\left(\frac{1}{(x^{2}+3x)^{24}+2}\right)^{-4/5} \\[1ex] \amp =\frac{1}{5}\left((x^{2}+3x)^{24}+2\right)^{4/5}, \end{split} \end{equation*}

and

\begin{equation*} h'(g(x)) = h'\left((x^{2}+3x)^{24}\right) = -\frac{1}{\left((x^{2}+3x)^{24}+2\right)^{2}} \cdot \end{equation*}

We can finally put together the derivative we are seeking,

\begin{equation*} \begin{split} \frac{d}{dx}\left((f \circ h \circ g)(x)\right) \amp = \frac{1}{5}\left((x^{2}+3x)^{24}+2\right)^{4/5} \cdot \left(-\frac{1}{\left((x^{2}+3x)^{24}+2\right)^{2}}\right) \cdot 24 \left(2x + 3\right)\left(x^{2}+3x\right)^{23} \\[1ex] \amp = -\frac{24(2x+3)(x^{2}+3x)^{23}((x^{2}+3x)^{24}+2)^{4/5}}{5((x^{2}+3x)^{24}+2)^{2}} \cdot \end{split} \end{equation*}
2. In a similar fashion, we find the first derivative of $\left(g \circ f \circ h\right)(x)$ to be

\begin{equation*} \begin{split} \frac{d}{dx}\left(g \circ f \circ h\right)(x) \amp = 24\left(2\sqrt[5]{\frac{1}{x+2}} + 3\right)\left(\sqrt[5]{\frac{1}{x+2}}^{2} + 3 \sqrt[5]{\frac{1}{x+2}}\right)^{23} \cdot \frac{1}{5}\left(\frac{1}{x+2}\right)^{-4/5} \cdot \left(-\frac{1}{(x+2)^{2}}\right) \\[1ex] \amp = 24\left(2\sqrt[5]{\frac{1}{x+2}} + 3\right)\left(\sqrt[5]{\frac{1}{x+2}}^{2} + 3 \sqrt[5]{\frac{1}{x+2}}\right)^{23} \cdot \frac{1}{5}\left(x+2 \right)^{4/5} \cdot \left(-\frac{1}{(x+2)^{2}}\right) \\[1ex] \amp = \frac{24\left(2\sqrt[5]{\frac{1}{x+2}} + 3\right)\left(\sqrt[5]{\frac{1}{x+2}}^{2} + 3 \sqrt[5]{\frac{1}{x+2}}\right)^{23} \left(x+2 \right)^{4/5}}{5(x+2)^{2}} \end{split} \end{equation*}
###### Example4.52. Rate of Change Application.

The membership of The Fitness Centre, which opened a few years ago, is approximated by the function

\begin{equation*} N(t)=100(64+4t)^{2/3} \ \ \ \ 0 \leq t \leq 52 \end{equation*}

where $N(t)$ gives the number of members at the beginning of week $t\text{.}$

1. Find $N'(t)\text{.}$

2. How fast was the centre's membership increasing initially ($t=0$)?

3. How fast was the membership increasing at the beginning of the $40$th week?

4. What was the membership when the centre first opened? At the beginning of the $40$th week?

Solution
1. Using the general Power Rule, we obtain

\begin{equation*} \begin{split} N'(t) \amp = \frac{d}{dt}\left[100(64+4t)^{2/3}\right] \\ \amp = 100 \frac{d}{dt} (64+4t)^{2/3} \\ \amp = \frac{200}{3}(64+4t)^{-1/3}\frac{d}{dt}(64+4t)\\ \amp = \frac{200}{3}(64+4t)^{-1/3}(4) \\ \amp = \frac{800}{3(64+4t)^{1/3}} \end{split} \end{equation*}
2. The rate at which the membership was increasing when the centre first opened is given by

\begin{equation*} N'(0) = \frac{800}{3(64)^{1/3}} \approx 66.7\text{,} \end{equation*}

or approximately $67$ people per week.

3. The rate at which the membership was increasing at the beginning of the $40$th week is given by

\begin{equation*} N'(40) = \frac{800}{3(64+160)^{1/3}} \approx 43.9\text{,} \end{equation*}

or approximately $44$ people per week.

4. The membership when the centre first opened is given by

\begin{equation*} N(0)= 100(64)^{2/3} = 100(16)\text{,} \end{equation*}

or approximately $1600$ people. The membership at the beginning of the $40$th week is given by

\begin{equation*} N(40) = 100(64+160)^{2/3} \approx 3688.3\text{,} \end{equation*}

or approximately $3688$ people.

##### Exercises for Section 4.4.

Find the derivatives of the functions. For extra practice, and to check your answers, do some of these in more than one way if possible.

1. $\ds f(x) = x^4-3x^3+(1/2)x^2+7x-\pi$

$\ds 4x^3-9x^2+x+7$
Solution
\begin{aligned}f'(x) \amp = \diff{}{x} \left(x^4-3x^3+(1/2)x^2+7x-\pi\right) \\ \amp = 4x^3 - 9x^2 + x + 7 \end{aligned}
2. $\ds f(x) = x^3-2x^2+4\sqrt{x}$

$\ds 3x^2-4x+2/\sqrt{x}$
Solution
\begin{aligned}f'(x) \amp = \diff{}{x} \left(x^3 - 2x^2 + 4\sqrt{x}\right) \\ \amp = 3x^2 - 4x + \frac{2}{\sqrt{x}} \end{aligned}
3. $\ds g(t) = (t^2+1)^3$

$\ds 6t(t^2+1)^2$
Solution
\begin{aligned}g'(t) \amp = \diff{}{t} \left(t^2+1\right)^3 \\ \amp = 3\left(t^2+1\right)^2 \diff{}{t} \left(t^2+1\right)\\ \amp = 3\left(t^2+1\right)^2 (2t) \\ \amp = 6t\left(t^2+1\right)^2 \end{aligned}
4. $\ds h(x) = x\sqrt{169-x^2}$

$\ds \sqrt{169-x^2}-x^2/\sqrt{169-x^2}$
Solution
\begin{aligned}h'(x) \amp = \diff{}{x} x\sqrt{169-x^2} \\ \amp = \sqrt{169-x^2} \diff{}{x} (x) + x \diff{}{x} \sqrt{169-x^2}\\ \amp = \sqrt{169-x^2} + x \frac{1}{2\sqrt{169-x^2}} \diff{}{x} \left(169-x^2\right) \\ \amp = \sqrt{169-x^2} + x \frac{1}{2\sqrt{169-x^2}} (-2x) \\ \amp = \sqrt{169-x^2} - \frac{x^2}{\sqrt{169-x^2}} \end{aligned}
5. $\ds f(t) = (t^2-4t+5)\sqrt{25-t^2}$

$\ds (2t-4)\sqrt{25-t^2}-$$(t^2-4t+5)t/\sqrt{25-t^2}$
Solution
\begin{aligned}f'(t) \amp = \diff{}{t} (t^2-4t+5)\sqrt{25-t^2} \\ \amp = \sqrt{25-t^2} \diff{}{t} (t^2-4t+5) + (t^2-4t+5) \diff{}{t} \sqrt{25-t^2} \\ \amp = \sqrt{25-t^2} (2t-4) + (t^2-4t+5) \frac{1}{2\sqrt{25-t^2}} \diff{}{t} \left(25-t^2\right) \\ \amp = \sqrt{25-t^2} (2t-4) + (t^2-4t+5) \frac{1}{2\sqrt{25-t^2}} (-2t)\\ \amp = \sqrt{25-t^2} (2t-4) - (t^2-4t+5) \frac{t}{\sqrt{25-t^2}} \end{aligned}
6. $\ds f(x) = \sqrt{r^2-x^2}\text{,}$ $r$ is a constant

$\ds -x/\sqrt{r^2-x^2}$
Solution
\begin{aligned}f'(x) \amp = \diff{}{x} \sqrt{r^2-x^2}\\ \amp = \frac{1}{2\sqrt{r^2-x^2}} \diff{}{x} \left(r^2-x^2\right) \\ \amp = \frac{1}{2\sqrt{r^2-x^2}} (-2x) \\ \amp = -\frac{x}{\sqrt{r^2-x^2}} \end{aligned}
7. $\ds h(s) = \sqrt{1+s^4}$

$\ds 2s^3/\sqrt{1+s^4}$
Solution
\begin{aligned}h'(s) \amp = \diff{}{s} \sqrt{1+s^4} \\ \amp = \frac{1}{2\sqrt{1+s^4}} \diff{}{s} \left(1+s^4\right) \\ \amp = \frac{1}{2\sqrt{1+s^4}} (4s^3) \\ \amp =\frac{2s^3}{\sqrt{1+s^4}} \end{aligned}
8. $\ds g(x) = {1\over\sqrt{5-\sqrt{x}}}$

$\ds{1\over 4\sqrt{x}(5-\sqrt{x})^{3/2}}$
Solution

We first rewrite $g(x) = \dfrac{1}{\sqrt{5-\sqrt{x}}}$ as $\left(5-x^{1/2}\right)^{-1/2}\text{.}$ We now take the derivative, applying the chain rule as needed.

\begin{equation*} \begin{split} g'(x) \amp = \diff{}{x} \left(5-x^{1/2}\right)^{-1/2} \\ \amp = -\frac{1}{2}\left(5-x^{1/2}\right)^{-3/2}\diff{ }{x} \left(5-x^{-1/2}\right) \\ \amp = \frac{1}{2}\left(5-x^{1/2}\right)^{-3/2}\left(-\frac{1}{2}x^{-1/2}\right)\\ \amp =\frac{1}{4\sqrt{x}\left(5-\sqrt{x}\right)^{3/2}}\cdot\end{split} \end{equation*}
9. $\ds f(x) = (1+3x)^2$

$\ds 6+18x$
Solution
\begin{aligned}\diff{f}{x} \amp = \diff{ }{x} \left(1+3x\right)^{2} \amp\\ \amp = 2\left(1+3x\right)\cdot \diff{ }{x}\left(1+3x\right)\amp\\ \amp = 2\left(1+3x\right)(3) \amp\\ \amp = 6\left(1+3x\right).\amp \end{aligned}
10. $\ds f(x) = {(x^2+x+1)\over(1-x)}$

$\ds {2 x + 1\over1 - x }+{x^2 + x + 1\over(1 - x)^2}$
Solution
\begin{aligned}\diff{f}{x} \amp = \diff{ }{x} \left(\frac{x^{2}+x+1}{1-x}\right)\amp\\ \amp = \diff{ }{x} (x^{2}+x+1)(1-x)^{-1}\amp\\ \amp = (1-x)^{-1}\diff{ }{x}\left(x^{2}+x+1\right) + (x^{2}+x+1)\diff{ }{x}\left(1-x\right)^{-1}\amp \\ \amp = (1-x)^{-1}\left(2x+1\right) + (x^{2}+x+1)(-1)(1-x)^{-2} \cdot \diff{ }{x}\left(1-x\right)\amp \\ \amp = (1-x)^{-1}\left(2x+1\right) + (x^{2}+x+1)(-1)(1-x)^{-2} \cdot \left(-1\right)\amp \\ \amp = \frac{2x+1}{1-x} + \frac{x^{2}+x+1}{(1-x)^{2}} \cdot\amp \end{aligned}
11. $\ds h(x) = {\sqrt{25-x^2}\over x}$

$\ds -1/\sqrt{25-x^2}-\sqrt{25-x^2}/x^2$
Solution
\begin{aligned}h'(x) \amp = \diff{}{x} \frac{\sqrt{25-x^2}}{x} \\ \amp = \frac{x\diff{}{x} \sqrt{25-x^2} - \sqrt{25-x^2} \diff{}{x} (x) }{x^2} \\ \amp = \frac{x\left(2\sqrt{25-x^2}\right)^{-1}\diff{}{x} (25-x^2) - \sqrt{25-x^2}}{x^2}\\ \amp = \frac{x\left(2\sqrt{25-x^2}\right)^{-1}(-2x) - \sqrt{25-x^2}}{x^2}\\ \amp = -\frac{x^2\left(\sqrt{25-x^2}\right)^{-1} + \sqrt{25-x^2}}{x^2} \end{aligned}
12. $\ds f(t) = \sqrt{{169\over t}-t}$

$\ds{1\over2}\left({-169\over t^2}-1\right)\Big/\sqrt{{169\over t}-t}$
Solution
\begin{aligned}f'(t) \amp = \diff{}{t} \sqrt{\frac{169}{t} - t}\\ \amp = \frac{1}{2\sqrt{\frac{169}{t} - t}} \diff{}{t} \left(\frac{169}{t} - t\right) \\ \amp = \frac{1}{2\sqrt{\frac{169}{t} - t}} \left(\frac{-169}{t^2} - 1\right) \end{aligned}
13. $\ds f(x) = \sqrt{x^3-x^2-(1/x)}$

$\ds{3x^2-2x+1/x^2\over 2\sqrt{x^3-x^2-(1/x)}}$
Solution
\begin{aligned}f'(x) \amp = \diff{}{x} \sqrt{x^3-x^2-(1/x)}\\ \amp = \frac{1}{2\sqrt{x^3-x^2-(1/x)}} \diff{}{x} \left(x^3-x^2-(1/x)\right) \\ \amp = \frac{1}{2\sqrt{x^3-x^2-(1/x)}} \left( 3x^2 - 2x + \frac{1}{x^2}\right) \end{aligned}
14. $\ds f(x) = 100/(100-x^2)^{3/2}$

$\ds{300 x \over(100-x^2)^{5/2}}$
Solution
\begin{aligned}f'(x) \amp = \diff{}{x} 100 \left(100-x^2\right)^{-3/2} \\ \amp = 100 \left(-\frac{3}{2} \left(100-x^2\right)^{-5/2}\right) \diff{}{x} (100-x^2) \\ \amp = -150 \left(100-x^2\right)^{-5/2} (-2x) \\ \amp = 300x \left(100-x^2\right)^{-5/2} \end{aligned}
15. $\ds g(t) = {\root 3 \of{t+t^3}}$

$\ds{ 1 + 3 t^2\over3(t+t^3)^{2/3}}$
Solution
\begin{aligned}g'(t) \amp = \diff{}{t} \left(t+t^3\right)^{1/3} \\ \amp = \frac{1}{3} \left(t+t^3\right)^{-2/3} \diff{}{t} (t+t^3) \\ \amp = \frac{1}{3} \left(t+t^3\right)^{-2/3} (1 + 3t^2) \\ \amp = \frac{1+3t^2}{3(t+t^3)^{2/3}} \end{aligned}
16. $\ds h(x) = \sqrt{(x^2+1)^2+\sqrt{1+(x^2+1)^2}}$

$\ds \left(4x(x^2+1)+{4x^3+4x\over2\sqrt{1+(x^2+1)^2}}\right)\Big/$$2\sqrt{(x^2+1)^2+\sqrt{1+(x^2+1)^2}}$
Solution
\begin{aligned}h'(x) \amp = \diff{}{x} \sqrt{(x^2+1)^2 + \sqrt{1+(x^2+1)^2}} \\ \amp = \frac{1}{2\sqrt{(x^2+1)^2 + \sqrt{1+(x^2+1)^2}}} \diff{}{x} \left((x^2+1)^2 + \sqrt{1+(x^2+1)^2}\right) \\ \amp = \frac{1}{2\sqrt{(x^2+1)^2 + \sqrt{1+(x^2+1)^2}}} \left(2(x^2+1)(2x) + \diff{}{x} \sqrt{1+(x^2+1)^2} \right) \\ \amp = \frac{1}{2\sqrt{(x^2+1)^2 + \sqrt{1+(x^2+1)^2}}} \left(2(x^2+1)(2x) + \frac{1}{2\sqrt{1+(x^2+1)^2}} \diff{}{x} (1+(x^2+1)^2) \right) \\ \amp = \frac{1}{2\sqrt{(x^2+1)^2 + \sqrt{1+(x^2+1)^2}}} \left(2(x^2+1)(2x) + \frac{1}{2\sqrt{1+(x^2+1)^2}} 2(x^2+1)(2x) \right) \\ \amp = \frac{1}{2\sqrt{(x^2+1)^2 + \sqrt{1+(x^2+1)^2}}} \left(4x(x^2+1) + \frac{4x(x^2+1)}{2\sqrt{1+(x^2+1)^2}}\right) \end{aligned}
17. $\ds f(x) = (x+8)^5$

$\ds 5(x+8)^4$
Solution
\begin{aligned}f'(x) \amp = \diff{}{x} (x+8)^5 \\ \amp = 5(x+8)^4 \diff{}{x} (x+8) \\ \amp = 5(x+8)^4 (1) \\ \amp = 5(x+8)^4 \end{aligned}

Find the derivatives of the functions. For extra practice, and to check your answers, do some of these in more than one way if possible.

1. $\ds f(t) = (4-t)^3$

$\ds -3(4-t)^2$
Solution
\begin{aligned}f'(t) \amp = \diff{}{t} (4-t)^3 \\ \amp = 3(4-t)^2 \diff{}{t} (4-t) \\ \amp = 3(4-t)^2 (-1) \\ \amp = -3(4-t)^2 \end{aligned}
2. $\ds f(x) = (x^2+5)^3$

$\ds 6x(x^2+5)^2$
Solution
\begin{aligned}f'(x) \amp = \diff{}{x} (x^2+5)^3 \\ \amp = 3(x^2+5)^2 \diff{}{x} (x^2 + 5) \\ \amp = 3(x^2+5)^2 (2x) \\ \amp = 6x (x^2+5)^2 \end{aligned}
3. $\ds f(x) = (6-2x^2)^3$

$\ds -12x(6-2x^2)^2$
Solution
\begin{aligned}f'(x) \amp = \diff{}{x} (6-2x^2)^3 \\ \amp = 3(6-2x^2)^2 \diff{}{x} (6-2x^2) \\ \amp = 3(6-2x)^2 (-4x) \\ \amp = -12x (6-2x)^2 \end{aligned}
4. $\ds g(x) = (1-4x^3)^{-2}$

$\ds 24x^2(1-4x^3)^{-3}$
Solution
\begin{aligned}g'(x) \amp = \diff{}{x} (1-4x^3)^{-2} \\ \amp = -2 (1-4x^3)^{-3} \diff{}{x} (1-4x^3) \\ \amp = -2 (1-4x^3)^{-3} (-12x^2) \\ \amp = 24x^2 (1-4x^3)^{-3} \end{aligned}
5. $\ds f(x) = 5(x+1-1/x)$

$\ds 5+5/x^2$
Solution
\begin{aligned}f'(x) \amp = \diff{}{x} 5 \left(x + 1 - \frac{1}{x} \right) \\ \amp = 5 \diff{}{x} \left(x + 1 - \frac{1}{x} \right) \\ \amp = 5\left( \diff{}{x} (x) + \diff{}{x} (1) - \diff{}{x} \frac{1}{x} \right)\\ \amp = 5\left(1 + 0 + \frac{1}{x^2}\right) \\ \amp = 5 + \frac{5}{x^2} \end{aligned}
6. $\ds f(t) = 4(2t^2-t+3)^{-2}$

$\ds -8(4t-1)(2t^2-t+3)^{-3}$
Solution
\begin{aligned}f'(x) \amp = \diff{}{x} 4 (2t^2-t+3)^{-2} \\ \amp = 4 \diff{}{x} (2t^2-t+3)^{-2} \\ \amp = 4 (-2 (2t^2-t+3)^{-3}) \diff{}{x} (2t^2-t+3) \\ \amp = -8(2t^2-t+3)^{-3} (4t - 1) \\ \amp = \frac{8(1-4t)}{(2t^2-t+3)^3} \end{aligned}
7. $\ds h(s) = {1\over 1+1/s}$

$\ds 1/(s+1)^2$
Solution
\begin{aligned}h'(s) \amp = \diff{}{s}\left( \frac{1}{1+1/s}\right) \\ \amp = \diff{}{x} \left(1+\frac{1}{s}\right)^{-1} \\ \amp = - \left(1+\frac{1}{s}\right)^{-2} \diff{}{s} \left(1+\frac{1}{s}\right) \\ \amp = - \left(1+\frac{1}{s}\right)^{-2} \left(-\frac{1}{s^2}\right) \\ \amp = \frac{1}{s^2\left(1+\frac{1}{s}\right)^2} \\ \amp = \frac{1}{(s+1)^2} \end{aligned}
8. $\ds f(x) = {-3\over 4x^2-2x+1}$

$\ds 3(8x-2)/(4x^2-2x+1)^2$
Solution
\begin{aligned}f'(x) \amp = \diff{}{x} \left(\frac{-3}{4x^2-2x+1}\right) \\ \amp = -3 \diff{}{x} \left(4x^2-2x+1\right)^{-1} \\ \amp = -3 (-1) \left(4x^2-2x+1\right)^{-2} \diff{}{x} \left(4x^2-2x+1\right)\\ \amp = 3 \left(4x^2-2x+1\right)^{-2} \left(8x - 2\right) \\ \amp = \frac{3(8x-2)}{\left(4x^2-2x+1\right)^{2}} \end{aligned}
9. $\ds f(x) = (x^2+1)(5-2x)/2$

$\ds -3x^2+5x-1$
Solution
\begin{aligned}f'(x) \amp = \diff{}{x} (x^2+1)(5-2x)/2 \\ \amp = \frac{1}{2} \diff{}{x} (x^2+1)(5x-2) \\ \amp = \frac{1}{2} \left( (5-2x) \diff{}{x} (x^2+1) + (x^2+1) \diff{}{x} (5-2x) \right) \\ \amp = \frac{1}{2} \left((5-2x) (2x) + (x^2+1) (-2) \right) \\ \amp = x(5-2x) - x^2 + 1 \\ \amp = -3x^3+5x+1 \end{aligned}
10. $\ds g(t) = (3t^2+1)(2t-4)^3$

$\ds 6t(2t-4)^3+6(3t^2+1)(2t-4)^2$
Solution
\begin{aligned}g'(t) \amp = \diff{}{t} (3t^2+1)(2t-4)^3 \\ \amp = (3t^2+1) \diff{}{t} (2t-4)^3 + (2t-4)^3 \diff{}{t} (3t^2+1) \\ \amp = (3t^2+1) \left(3(2t-4)^2 \diff{}{t} (2t-4) \right) + (2t-4)^3 (6t) \\ \amp = (3t^2+1) \left(3(2t-4)^2 (2) \right) + (2t-4)^3 (6t) \\ \amp = 6(3t^2+1)(2t-4)^2 + 6t (2t-4)^3 \end{aligned}
11. $\ds f(x) = {x+1\over x-1}$

$\ds -2/(x-1)^2$
Solution
\begin{aligned}f'(x) \amp = \diff{}{x} \left(\frac{x+1}{x-1}\right) \\ \amp = \frac{(x-1) \diff{}{x} (x+1) - (x+1) \diff{}{x} (x-1)}{(x-1)^2} \\ \amp = \frac{(x-1) (1) - (x+1)(1)}{(x-1)^2} \\ \amp = \frac{-2}{(x-1)^2} \end{aligned}
12. $\ds h(s) = {s^2-1\over s^2+1}$

$\ds 4s/(s^2+1)^2$
Solution
\begin{aligned}h'(s) \amp = \diff{}{s}\left( \frac{s^2-1}{s^2+1}\right) \\ \amp = \frac{(s^2+1) \diff{}{s} (s^2-1) - (s^2-1) \diff{}{s} (s^2+1)}{(s^2+1)^2} \\ \amp = \frac{(s^2+1)(2s) - (s^2-1)(2s)}{(s^2+1)^2}\\ \amp = \frac{4s}{(s^2+1)^2} \end{aligned}
13. $\ds h(t) = {(t-1)(t-2)\over t-3}$

$\ds (t^2-6t+7)/(t-3)^2$
Solution
\begin{aligned}h'(t) \amp = \diff{}{t}\left( \frac{(t-1)(t-2)}{t-3}\right) \\ \amp = \frac{(t-3) \diff{}{t} (t-1)(t-2) - (t-1)(t-2) \diff{}{t} (t-3) }{(t-3)^2} \\ \amp = \frac{(t-3)\left((t-1) + (t-2)\right) - (t-1)(t-2) (1)}{(t-3)^2} \\ \amp = \frac{(t-3)\left(2t-3\right) - (t-1)(t-2)}{(t-3)^2}\\ \amp = \frac{t^2-6t+7}{(t-3)^2} \end{aligned}
14. $\ds g(x) = {2x^{-1}-x^{-2}\over 3x^{-1}-4x^{-2}}$

$\ds -5/(3x-4)^2$
Solution

First, we calculate

\begin{equation*} \diff{}{x} \left(\frac{2}{x} - \frac{1}{x^2}\right) = -\frac{2}{x^2} + \frac{2}{x^3} = \frac{2-2x}{x^3}\text{,} \end{equation*}

and

\begin{equation*} \diff{}{x} \left(\frac{3}{x} - \frac{4}{x^2} \right) = -\frac{3}{x^2} + \frac{8}{x^3} = \frac{8-3x}{x^3}\text{.} \end{equation*}

Therefore, by the quotient rule:

\begin{equation*} \begin{split} g'(x) \amp = \diff{}{x}\left( \frac{\frac{2}{x} - \frac{1}{x^2}}{\frac{3}{x} - \frac{4}{x^2}}\right) \\ \amp = \frac{\left(\frac{3}{x} - \frac{4}{x^2}\right)\frac{2-2x}{x^3} - \left(\frac{2}{x} - \frac{1}{x^2}\right) \frac{8-3x}{x^3}}{\left(\frac{3}{x} - \frac{4}{x^2} \right)^2}\\ \amp = \frac{\left(\frac{3x-4}{x^2}\right)\frac{2-2x}{x^3} - \left(\frac{2x-1}{x^2}\right) \frac{8-3x}{x^3}}{\frac{(3x-4)^2}{x^4}}\\ \amp =\frac{\frac{(3x-4)(2-2x)}{x}- \frac{(2x-1)(8-3x)}{x}}{(3x-4)^2}\\ \amp = \frac{-5}{(3x-4)^2} \end{split} \end{equation*}
15. $\ds f(x) = 3(x^2+1)(2x^2-1)(2x+3)$

$\ds 60x^4+72x^3+18x^2+18x-6$
Solution
\begin{aligned}f'(x) \amp = \diff{}{x} 3 (x^2+1)(2x^2-1)(2x+3) \\ \amp = 3 (2x) (2x^2-1)(2x+3) + 3(x^2+1)(4x)(2x+3) + 3(x^2+1)(2x^2-1)(2) \\ \amp = 60 x^4 + 72 x^3 + 18 x^2 + 18 x - 6 \end{aligned}
16. $\ds f(x) = {1\over (2x+1)(x-3)}$

$\ds (5-4x)/((2x+1)^2(x-3)^2)$
Solution
\begin{aligned}f'(x) \amp = \diff{}{x} \left(\frac{1}{(2x+1)(x-3)}\right) \\ \amp =\diff{}{x} (2x+1)^{-1} (x-3)^{-1} \\ \amp = (2x+1)^{-1} \left(-(x-3)^{-2}(1)\right) + (x-3)^{-1} \left(-(2x+1)^{-2}(2) \right) \\ \amp =\frac{-1}{(2x+1)(x-3)^2} - \frac{2} {(x-3)(2x+1)^2} \end{aligned}
17. $\ds f(s) = ((2s+1)^{-1}+3)^{-1}$

$\ds 1/(2(2+3s)^2)$
Solution
\begin{aligned}f'(s) \amp = \diff{}{s} \bigl((2s+1)^{-1}+3\bigr)^{-1} \\ \amp = (-1) \bigl((2s+1)^{-1}+3\bigr)^{-2} \diff{}{s} \bigl((2s+1)^{-1}+3\bigr) \\ \amp = -\bigl((2s+1)^{-1}+3\bigr)^{-2} (- (2s+1)^{-2} (2) ) \\ \amp = \frac{2(2s+1)^{-2}}{((2s+1)^{-1}+3)^2}\\ \amp = \frac{1}{2 (2+3s)^2} \end{aligned}
18. $\ds h(x) = (2x+1)^3(x^2+1)^2$

$\ds 56x^6+72x^5+110x^4+100x^3+60x^2+28x+6$
Solution
\begin{aligned}h'(x) \amp = \diff{}{x} (2x+1)^3(x^2+1)^2 \\ \amp = (2x+1)^3 \diff{}{x} (x^2+1)^2 + (x^2 + 1)^2 \diff{}{x} (2x+1)^3 \\ \amp = (2x+1)^2 \left( 2(x^2+1) (2x) \right) + (x^2+1)^2 \left(3(2x+1)^2 (2) \right) \\ \amp = 4x (2x+1)^2(x^2+1) + 6(x^2+1)^2(2x+1)^2 \end{aligned}

Find an equation for the tangent line to $\ds f(x) = (x-2)^{1/3}/(x^3 + 4x - 1)^2$ at $x=1\text{.}$

$y=23x/96-29/96$
Solution

Let $f(x) = \frac{(x-2)^{1/3}}{(x^3+4x-1)^2}\text{.}$ We first compute $f'(x)\text{:}$

\begin{equation*} \begin{split} f'(x) \amp = \diff{}{x} \frac{(x-2)^{1/3}}{(x^3+4x-1)^2}\\ \amp = \frac{(x^3+4x-1)^2 \diff{}{x} (x-2)^{1/3} - (x-2)^{1/3} \diff{}{x} (x^3+4x-1)^2}{(x^3-4x+1)^4} \\ \amp = \frac{(x^3+4x-1)^2 \left( \frac{1}{3}(x-2)^{-2/3} \right) - (x-2)^{1/3} \left(2 (x^3+4x-1) (3x^2+4)\right)}{(x^3-4x+1)^4} \end{split} \end{equation*}

Therefore, when $x=1\text{,}$ the slope of the tangent line is

\begin{equation*} f'(1) = \frac{4^2 \left(\frac{1}{3}\right) - (-1) (2) (4) (7)}{4^4} = \frac{23}{96}\text{.} \end{equation*}

Since the point of tangency is $(1,f(1)) = (1,-1/16)\text{,}$ an equation of the tangent line is

\begin{equation*} y + \frac{1}{16} = \frac{23}{96} \left(x - 1 \right)\text{.} \end{equation*}

Find an equation for the tangent line to $\ds y=9x^{-2}$ at $(3,1)\text{.}$

$y=3-2x/3$
Solution

Let $f(x) = 9x^{-2}\text{.}$ We first compute the derivative $f'(x)\text{:}$

\begin{equation*} \begin{split} f'(x) \amp = \diff{}{x} (9x^{-2}) \\ \amp = 9 (-2 x^{-3} ) \\ \amp = -18 x^{-3} \end{split} \end{equation*}

Therefore, when $x=3\text{,}$ the slope of the tangent line is

\begin{equation*} f'(3) = -\frac{18}{3^3} = -\frac{2}{3}\text{.} \end{equation*}

The point of tangency is $(3,1)\text{,}$ and so an equation of the tangent line is

\begin{equation*} y - 1 = - \frac{2}{3} (x-3)\text{.} \end{equation*}

Find an equation for the tangent line to $\ds (x^2-4x+5)\sqrt{25-x^2}$ at $(3,8)\text{.}$

$y=13x/2-23/2$
Solution

We use the product rule and subsequently the chain rule to compute $f'(x)\text{:}$

\begin{equation*} \begin{split} f'(x) \amp = \diff{}{x} (x^2-4x+5) \sqrt{25-x^2} \\ \amp = \sqrt{25-x^2} \diff{}{x} (x^2-4x+5) + (x^2-4x+5) \diff{}{x} \sqrt{25-x^2} \\ \amp = \sqrt{25-x^2}(2x-4) + (x^2-4x+5) \left(\frac{1}{2} \left(25-x^2\right)^{-1/2} \cdot \diff{}{x} \left(25-x^2\right)\right) \\ \amp = \sqrt{25-x^2}(2x-4) + (x^2-4x+5) \left(\frac{1}{2} \left(25-x^2\right)^{-1/2} \cdot (-2x)\right) \\ \amp =\sqrt{25-x^2}(2x-4) - \frac{x(x^2-4x+5)}{\sqrt{25-x^2}}\\ \amp = \frac{-3x^3+8x^2+4x-100}{\sqrt{25-x^2}} \end{split} \end{equation*}

So, at the point $(3,8)\text{,}$ the slope $m$ of the tangent line to $f$ is

\begin{equation*} m = \frac{-3(3)^3+8(3)^2+4(3)-100}{\sqrt{25-(3)^2}} = \frac{13}{2}\text{.} \end{equation*}

Therefore, we have

\begin{equation*} y-8 = \frac{13}{2}(x-3)\text{,} \end{equation*}

or,

\begin{equation*} y = \frac{13}{2} x -\frac{23}{2}\text{.} \end{equation*}

Find an equation for the tangent line to $\ds \ds{(x^2+x+1)\over(1-x)}$ at $(2,-7)\text{.}$

$y=2x-11$
Solution

Let $f(x) = \dfrac{(x^2+x+1)}{(1-x)}\text{.}$ Now compute the derivative $f'(x)\text{:}$

\begin{equation*} \begin{split} f'(x) \amp = \diff{}{x} \dfrac{(x^2+x+1)}{(1-x)}\\ \amp = \dfrac{(1-x)\diff{}{x} (x^2+x+1) - (x^2+x+1)\diff{}{x} (1-x) }{(1-x)^2}\\ \amp = \dfrac{(1-x)(2x+1) + (x^2+x+1) }{(1-x)^2}\\ \amp = \dfrac{(1-x)(2x+1) + (x^2+x+1) }{(1-x)^2}\\ \amp = -\dfrac{x^2 - 2 x - 2 }{(1-x)^2} \end{split} \end{equation*}

Therefore, when $x=2\text{,}$ the slope of the tangent line is

\begin{equation*} f'(2) = - \frac{4-4-2}{1} = 2\text{.} \end{equation*}

The point of tangency is $(2,-7)\text{,}$ and so an equation of the tangent line is

\begin{equation*} y + 7 = 2(x-2)\text{.} \end{equation*}

Find an equation for the tangent line to $\ds \sqrt{(x^2+1)^2+\sqrt{1+(x^2+1)^2}}$ at $\ds (1,\sqrt{4+\sqrt{5}})\text{.}$

$\ds y={20+2\sqrt5\over5\sqrt{4+\sqrt5}}\,x+{3\sqrt5\over5\sqrt{4+\sqrt5}}$
Solution

Let $f(x) = \sqrt{(x^2+1)^2 + \sqrt{1+(x^2+1)^2}}\text{.}$ Then:

\begin{equation*} \begin{split} f'(x) \amp = \diff{}{x} \sqrt{(x^2+1)^2 + \sqrt{1+(x^2+1)^2}}\\ \amp = \frac{1}{2 \sqrt{(x^2+1)^2 + \sqrt{1+(x^2+1)^2}}} \diff{}{x} \left((x^2+1)^2 + \sqrt{1+(x^2+1)^2}\right) \\ \amp = \frac{1}{2 \sqrt{(x^2+1)^2 + \sqrt{1+(x^2+1)^2}}} \left(2(x^2+1)(2x) + \frac{1}{2\sqrt{1+(x^2+1)^2}} \diff{}{x} \left(1+(x^2+1)^2\right)\right) \\ \amp = \frac{1}{2 \sqrt{(x^2+1)^2 + \sqrt{1+(x^2+1)^2}}} \left(2(x^2+1)(2x) + \frac{1}{2\sqrt{1+(x^2+1)^2}} (2(x^2+1) (2x))\right)\\ \amp = \frac{2x(x^2+1)}{\sqrt{(x^2+1)^2 + \sqrt{1+(x^2+1)^2}}} \left(1 + \frac{1}{2\sqrt{1+(x^2+1)^2}}\right) \end{split} \end{equation*}

Therefore, when $x=1\text{,}$ the slope of the tangent line is:

\begin{equation*} f'(1) = \frac{4}{\sqrt{4 + \sqrt{5}}} \left(1 + \frac{1}{2\sqrt{5}}\right)\text{.} \end{equation*}

Since the point of tangency is $\left(1, \sqrt{4+\sqrt{5}}\right)\text{,}$ an equation of the tangent line is

\begin{equation*} y - \sqrt{4+\sqrt{5}} = \frac{4}{\sqrt{4 + \sqrt{5}}} \left(1 + \frac{1}{2\sqrt{5}}\right) (x - 1)\text{.} \end{equation*}

Let $y=f(x)$ and $x=g(t)\text{.}$ If $g(1)=2\text{,}$ $f(2)=3\text{,}$ $g'(1)=4$ and $f'(2)=5\text{,}$ find the derivative of $f\circ g$ at 1.

$(f(g(1)))'=20$
Solution

Let $y = f(g(t))\text{.}$ We are given that $g(1)=2\text{,}$ $f(2)=3\text{,}$ $g'(1) = 4\text{,}$ and $f'(2)=5\text{,}$ and wish to find $\diff{y}{t}$ at $t=1\text{.}$ By the chain rule,

\begin{equation*} \diff{y}{t} = \diff{ }{t} f(g(t)) = f'(g(t)) \cdot g'(t)\text{.} \end{equation*}

So at $t=1\text{,}$

\begin{equation*} \diff{ }{t} f(g(1)) = f'(g(1)) \cdot g'(1) = f'(2) \cdot 4 = 5 \cdot 4 = 20\text{.} \end{equation*}

Use the graphical information of the functions $f$ and $g$ to find the derivative of the composition function and the given x-value.

1. $\ds (f \circ g)'(2)$

$0.5$
Solution

By the chain rule, $(f \circ g)'(2) = f'\left(g(2)\right) \cdot g'(2)\text{.}$ We can see from the graph that $g(2)= 4.5\text{.}$ Thus, we have $(f \circ g)'(2) = f'\left(4.5\right) \cdot g'(2)\text{.}$ At $x=2\text{,}$ the slope of $g(x)$ is

\begin{equation*} \frac{5-2}{3-(-3)} = \frac{1}{2}\text{,} \end{equation*}

and at $x=4.5\text{,}$ the slope of $f(x)$ is

\begin{equation*} \frac{3-1}{4-2} = 1\text{.} \end{equation*}

Therefore, $(f \circ g)'(2) = f'\left(g(2)\right) \cdot g'(2) = 1 \cdot \frac{1}{2} = \frac{1}{2}\text{.}$

2. $\ds (f \circ f)'(1)$

$1$
Solution

Similarly, we have $(f \circ f)'(1) = f'\left(f(1)\right)f'(1) = f'(0)f'(1) = (1)(1) = 1$ (here, we use the fact that $f'(0) = f'(1) = f'(4.5)$).

3. $\ds (g \circ f)'(3)$

$0.5$
Solution

$(g \circ f)'(3) = g'\left(f(3)\right)f'(3) = g'(2)f'(3)\text{.}$ Note that we already found $g'(2)$ in (a), and that $f'(3)= f'(4.5)\text{.}$ Therefore, $(g \circ f)'(3) = \frac{1}{2}\text{.}$

4. $\ds (f \circ g)'(3)$

DNE
Solution
$(f \circ g)'(3) = f'\left(g(3)\right)\cdot g'(3)\text{.}$ This derivative does not exist, since $g'(3)$ does not exist (there is a corner at $g(3)$).
5. $\ds (g \circ f)'(-2)$

DNE
Solution

$(g \circ f)'(-2) = g'\left(f(-2)\right)\cdot f'(-2)\text{.}$ Again, we find that the derivative does not exist, this time because $f'(-2)$ does not exist.

6. $\ds (g \circ f)'(-4)$

$-0.5$
Solution

$(g \circ f)'(-4) = g'\left(f(-4)\right)\cdot f'(-4) = g'(-1)f'(-4)\text{.}$ We see that $g'(-1) = g'(2) = \frac{1}{2}\text{,}$ and find

\begin{equation*} f'(-4) = \frac{-1-(-2)}{-4-(-3)} = -1\text{.} \end{equation*}

Therefore, $(g \circ f)'(-4) = -\frac{1}{2}\text{.}$

Express the derivative of $g(x)=x^2f(x^2)$ in terms of $f$ and the derivative of $f\text{.}$

$g'(x)=2x(f(x^2)+x^2f'(x^2))$
Solution

Let $g(x) = x^{2}f(x^{2})\text{.}$ Then,

\begin{equation*} \begin{split} \diff{g}{x} \amp = \diff{ }{x} x^{2}f(x^{2})\\ \amp = \left(\diff{ }{x} x^{2}\right)f(x^{2}) + x^{2}\left(\diff{ }{x} f(x^{2})\right) \\ \amp = 2xf(x^{2}) + x^{2}\left(f'(x^{2}) \diff{ }{x} x^{2}\right)\\ \amp = 2xf(x^{2}) + x^{2}\left(2xf'(x^{2})\right) \\ \amp = 2x \left(f(x^{2}) + x^{2}f'(x^{2})\right). \end{split} \end{equation*}

Let $F(x) = f(f(x))\text{.}$ Does it follow that $F'(x) = [f'(x)]^{2}\text{?}$ Hint

Let $f(x) = x^{2}\text{.}$

No.
Solution

By the Chain Rule,

\begin{equation*} \diff{F}{x} = \diff{}{x} f(f(x)) = f'(f(x)) \cdot f'(x)\text{.} \end{equation*}

Since, in general, $f'(f(x)) \neq f'(x)\text{,}$ the statement

\begin{equation*} \diff{F}{x} = \left[f'(x)\right]^2 \end{equation*}

is false. For a counterexample, take $f(x) = x^2\text{.}$ Then

\begin{equation*} F(x) = (x^2)^2 = x^4\text{.} \end{equation*}

Suppose $h= g \circ f\text{.}$ Does it follow that $h'=g' \circ f'\text{?}$ Hint

Let $f(x) = x$ and $g(x) = x^{2}\text{.}$

No.
Solution

Let $h(x) = g(f(x))\text{.}$ Therefore, by the Chain Rule,

\begin{equation*} h'(x) = g'(f(x)) \cdot f'(x) \neq g'(f'(x))\text{.} \end{equation*}

Therefore, the statement is false. For a counterexample, let $f(x) = x$ and $g(x) = x^2\text{.}$ Then

\begin{equation*} h'(x) = \diff{}{x} x^2 = 2x\text{.} \end{equation*}

However,

\begin{equation*} g'(f'(x)) = g'(1) = 2\text{.} \end{equation*}

The number of viewers of a television series introduced several years ago is approximated by the function

\begin{equation*} N(x)=(60+2x)^{2/3} \ \ \ 1 \leq x \leq 26 \end{equation*}

where $N(x)$ (measured in thousands) denotes the number of weekly viewers of the series in the $x$-th week. Find the rate of increase of the weekly audience at the end of week 2 and at the end of week 12. How many viewers were there in week 2? In week 24?

$\frac{1}{3}$ thousand/wk; 16 thousand; 22.7 thousand
Solution

We first differentiate $N(x)\text{:}$

\begin{equation*} N'(x) = \frac{2}{3} (60+2x)^{-1/3} (2) = \frac{4}{3} (60+2x)^{-1/3}, \ 1 \leq x \leq 26\text{.} \end{equation*}

We now compute $N'(2) = \frac{1}{3}$ and $N'(24) \approx 0.30\text{.}$ Therefore, at the end of the second week, the audience was increasing at a rate of about 0.33 thousand viewers per week. By the end of the twelfth week, the audience was increasing at a rate of about 0.3 thousand viewers per week.

We further calculate $N(2) = 16$ and $N(24) = 22.7\text{.}$ Therefore, at the end of the second week there were 16 thousand viewers. By week 24, there were 22.7 thousand viewers.

The registrar of a prominent city in Montreal estimates that the total student enrollment in the Continuing Education division will be given by

\begin{equation*} N(t) = -\frac{20,000}{\sqrt{1+0.2t}}+21,000 \end{equation*}

where $N(t)$ denotes the number of students enrolled in the division $t$ yr from now. Find an expression for $N'(t)\text{.}$ How fast is the student enrollment increasing currently? How fast will it be increasing 5 yr from now?

$N'(t) = \frac{2000}{(1+0.2t)^{3/2}}\text{;}$2000 students; 707 students
Solution

We differentiate $N(t)\text{:}$

\begin{equation*} \begin{split} N'(t) \amp = \diff{}{t} \left(-\frac{20,000}{\sqrt{1+0.2t}} + 21,000\right) \\ \amp = -20,000 \diff{}{t} \frac{1}{\sqrt{1+0.2t}} + \diff{}{t} 21,000 \\ \amp = -20,000 \left(\frac{-1}{2(1+0.2t)^{3/2}} \diff{}{t} (1+0.2t)\right)\\ \amp = 20,000\left(\frac{1}{2(1+0.2t)^{3/2}}\right) (0.2) \\ \amp = \frac{2000}{(1+0.2t)^{3/2}} \end{split} \end{equation*}

We now compute

\begin{equation*} N'(0) =\frac{2000}{(1)^{3/2}} = 2000\text{,} \end{equation*}

and

\begin{equation*} N'(5) = \frac{2000}{(1+1)^{3/2}} = 707.107\text{.} \end{equation*}

Therefore, the number of students is currently increasing at a rate of 2000 students per year. In 5 years, the number of students is expected to increase at a rate of about 707 students per year.

The president of a major housing construction firm claims that the number of construction jobs created is given by

\begin{equation*} N(x)=1.42x \end{equation*}

where $x$ denotes the number of housing starts. Suppose the number of housing starts in the next $t$ mo is expected to be

\begin{equation*} x(t)=\frac{7t^{2}+140t+700}{3t^{2}+80t+550} \end{equation*}

million units/year. Find an expression that gives the rate of change at which the number of construction jobs will be created $t$ mo from now. At what rate will construction jobs be created 1 yr from now?

$N'(t)=\frac{1.42(140t^{2}+3500t+21,000)}{(3t^{2}+80t+550)^{2}}\text{;}$ 31,312 jobs/yr
Solution

We define $h(t)$ to be the number of construction jobs created as a function of time. Then $h(t)$ is given by the composition,

\begin{equation*} h(t) = N(x(t))\text{,} \end{equation*}

where $t$ is in months. By the chain rule, the rate of change of job creation as a function of time is

\begin{equation*} \diff{h}{t} = N'(x(t)) \cdot x'(t) = 1.42 \left(\frac{140t^{2}+3500t+21,000)}{(3t^{2}+80t+550)^{2}}\right)\text{.} \end{equation*}

So, 1 year from now, we estimate the rate at which construction jobs are being created to be

\begin{equation*} h'(12) = 1.42 \left(\frac{140(12^{2})+3500(12)+21,000)}{(3(12)^{2}+80(12)+550)^{2}}\right) \approx 0.0313115\text{,} \end{equation*}

or approximately $31,312$ jobs/year.

Government economists of a developing country determined that the purchase of imported perfume is related to a proposed “luxury tax” by the formula

\begin{equation*} N(x)=\sqrt{10,000-40x-0.02x^{2}} \ \ \ \ 0 \leq x \leq 200 \end{equation*}

where $N(x)$ measured the percentage of normal consumption of perfume when a “luxury tax” of $x$% is imposed on it. Find the rate of change of $N(x)$ for taxes of 10, 100, and 150.

-20.6%; 28.9%; 38.6%.
Solution

We first find $N'(x)\text{:}$

\begin{equation*} \begin{split} N'(x) \amp = \diff{}{x} \sqrt{10,000 - 40 x - 0.02x^2}\\ \amp = \frac{1}{2\sqrt{10,000-40x-0.02x^2}} \diff{}{x} \left(10,000 - 40 x - 0.02x^2\right)\\ \amp = \frac{1}{2\sqrt{10,000-40x-0.02x^2}} \left(- 40 - 0.04x\right)\\ \amp = \frac{-2(10+0.01x)}{\sqrt{10,000-40x-0.02x^2}} \end{split} \end{equation*}

Therefore, when the luxury tax is set at 10%,

\begin{equation*} N'(10) = \frac{-2(10+0.1)}{\sqrt{10,000-400-2}} \approx -0.206\text{,} \end{equation*}

which means that the consumption of imported perfume is decreasing by approximately 20.6%.

Similarly, when the luxury tax is set at 100%,

\begin{equation*} N'(100) = \frac{-2(10+1)}{\sqrt{10,000 - 4000-200}} \approx -0.289\text{,} \end{equation*}

which means that the consumption of imported perfume is decreasing by approximately 28.9 %.

Finally, when the luxury tax is set at 150%,

\begin{equation*} N'(150) \approx -0.386\text{,} \end{equation*}

which means that the consumption of imported perfume is decreasing by approximately 38.6%.

The quantity demanded per month, $x\text{,}$ of a certain make of personal computer (PC) is related to the average unit price,$p$ (in dollars), of PCs by the equation

\begin{equation*} x = f(p) = \frac{100}{9}\sqrt{810,000 - p^{2}}\text{.} \end{equation*}

It is estimated that $t$ mo from now, the average price of a PC will be given by

\begin{equation*} p(t) = \frac{400}{1+\frac{\sqrt{t}}{8}}+200 \ \ \ 0\leq t \leq 60 \end{equation*}

dollars. Find the rate at which the quantity demanded per month of the PCs will be changed 16 months from now.

Approximately 19 computer/mo.
Solution

We want to find the rate of change of the quantity demanded $x$ with respect to time. We are given $x = f(p)$ where $p$ is a function of time. Therefore, we use the Chain Rule:

\begin{equation*} \diff{x}{t} = \diff{x}{p} \diff{p}{t}\text{.} \end{equation*}

When $t=16\text{,}$ we see that

\begin{equation*} p = \frac{400}{1+\frac{1}{2}} + 200 = \frac{1400}{3}\text{.} \end{equation*}

Therefore,

\begin{equation*} \diff{x}{t} \bigg\vert_{t=16} = \diff{x}{p} \bigg\vert_{p=1400/3} \diff{p}{t}\bigg\vert_{t=16}\text{.} \end{equation*}

We first compute:

\begin{equation*} \begin{split} \diff{x}{p} \amp = \diff{}{p} \left(\frac{100}{9}\sqrt{810,000 - p^2}\right)\\ \amp =\frac{100}{9} \frac{1}{2 \sqrt{810,000 - p^2}} \diff{}{p} \left(81,000 - p^2\right) \\ \amp = \frac{100}{9} \frac{1}{2\sqrt{810,000-p^2}} \left(-2p\right) \\ \amp = \frac{-100}{9 \sqrt{810,000-p^2}} \end{split} \end{equation*}

And so when $p=1400/3\text{,}$

\begin{equation*} \diff{x}{p}\bigg\vert_{p=1400/3} = \frac{-1400}{(9 \sqrt{533}}\text{.} \end{equation*}

Next, we need to find the derivative of $p$ with respect to $t\text{:}$

\begin{equation*} \begin{split} \diff{p}{t} \amp = \diff{}{t} \left(\frac{400}{1+\frac{\sqrt{t}}{8}} + 200\right)\\ \amp = 400 \diff{}{t} \left(1+\frac{\sqrt{t}}{8}\right)^{-1} \\ \amp = -400 \left(1+\frac{\sqrt{t}}{8}\right)^{-2} \diff{}{t} \left(1+\frac{\sqrt{t}}{8}\right)\\ \amp = -400 \left(1+\frac{\sqrt{t}}{8}\right)^{-2} \left(\frac{1}{16\sqrt{t}}\right)\\ \amp = \frac{-25}{\sqrt{t}\left(1+\frac{\sqrt{t}}{8}\right)^{2}} \end{split} \end{equation*}

And so, when $t=16\text{,}$

\begin{equation*} \diff{p}{t} \bigg\vert_{t=16} = \frac{-25}{9}\text{.} \end{equation*}

Altogether, we have

\begin{equation*} \diff{x}{t} \bigg\vert_{t=16} = \left(\frac{-1400}{(9 \sqrt{533}}\right)\left(\frac{-25}{9}\right) \approx 18.716\text{.} \end{equation*}

Therefore, after 16 months, the quantity demanded is increasing at a rate of about 19 computers per month.